According to new cartesian sign convention, Object distance, $u=-15 \mathrm{~cm}$ Focal length of the concave mirror, $f=-10 \mathrm{~cm}$ Height of the object, $h_O=2 \mathrm{~cm}$
According to mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \quad \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}-\frac{1}{-15}=\frac{1}{-10}+\frac{1}{15}$
or $\quad v=-30 \mathrm{~cm}$.
The image is formed at a distance of $30 \mathrm{~cm}$ from the mirror on the same side of the object. It is a real image.
Magnification of the mirror, $m=\frac{-v}{u}=\frac{h_I}{h_O}$
$\Rightarrow \frac{-(-30)}{-15}=\frac{h_I}{2} \Rightarrow h_I=-4 \mathrm{~cm}$.
Negative sign shows that image is inverted. The image is real, inverted, of size $4 \mathrm{~cm}$ at a distance of $30 \mathrm{~cm}$ in front of the mirror.